You have found the following ages (in years) of 4 gorillas. Those gorillas were randomly selected from the 29 gorillas at your local zoo: $ 1,\enspace 5,\enspace 5,\enspace 8$ Based on your sample, what is the average age of the gorillas? What is the standard deviation? You may round your answers to the nearest tenth.
Solution: Because we only have data for a small sample of the 29 gorillas, we are only able to estimate the population mean and standard deviation by finding the sample mean $({\overline{x}})$ and sample standard deviation $({s})$ To find the sample mean , add up the values of all $4$ samples and divide by $4$ $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\overline{x}} = \dfrac{1 + 5 + 5 + 8}{{4}} = {4.8\text{ years old}} $ Find the squared deviations from the mean for each sample. Since we don't know the population mean, estimate the mean by using the sample mean we just calculated {14.44} + {0.04} + {0.04} + {10.24}} {{4 - 1}} $ {s^2} = \dfrac{{24.76}}{{3}} = {8.25\text{ years}^2} $ As you might guess from the notation, the sample standard deviation $({s})$ is found by taking the square root of the sample variance $({s^2})$ ${s} = \sqrt{{s^2}}$ $ {s} = \sqrt{{8.25\text{ years}^2}} = {2.9\text{ years}} $ We can estimate that the average gorilla at the zoo is 4.8 years old. There is also a standard deviation of 2.9 years.